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Method to carry out the isobaric process

  1. Aug 27, 2012 #1
    hi ...

    These days i am trying to make sense of the various processes of thermodynamics and so far i have tried the isothermal process , and isochoric process (which is quite simple i think). Now i am trying to describe what i have understood by the isobaric process:

    According to the basic idea of pressure: it increases when the volume is reduced keeping temperature nearly constant, and also when the temperature is raised keeping the no of moles constant. With this idea in mind i thought of a way to carry out the process.

    If dQ amount of heat is supplied to the system in consideration, then its temperature increases by dT amount. Then the supply of heat is put off and the volume is allowed to increase by dV amount, and meanwhile the temperature is not allowed to fall below its initial value, in fact the source of heat is turned on as soon
    dV = nR dT/P. And again the same cycle is repeated so the pressure remains constant.

    So is it a method to carry out the process?
    Do you think it is making sense and taking the form of a quasi-static isobaric process?
    Or can you suggest a better method to carry out the isobaric process ?

    I am looking forward to your quick reply....
  2. jcsd
  3. Aug 28, 2012 #2
    I think you've got it right. For an ideal gas, as long as any change in temperature dT is accompanied by a change in volume dV = Nk/P dT, then pressure will remain constant. (Forgive me, but I really dislike the constant R, since R = k*A, where k is Boltzmann's constant and A is Avogadro's number, which is too chemistry-y for me.)

    One way I like to understand this process is by thinking of a bubble of gas surrounded by another gas at pressure P, or, alternatively, a gas inside a cylinder equipped with a piston exerting a constant force F on the gas. When the gas inside the bubble is heated causing a temperature change dT, the pressure inside the bubble must equilibrate with the pressure outside, so a change dV must occur. Likewise, when the gas inside the piston undergoes a temperature change dT, the volume must change so that the force exerted on the piston by the gas equilibrates with the force F on the piston.

    Maybe you could help me out with my as-yet unanswered question on adiabatic compression and joule expansion of gasses, which is a little further down in this board.
    Last edited: Aug 28, 2012
  4. Aug 30, 2012 #3
    I tried thinking this way of keeping pressure constant, but the number of molecules inside and outside the closed container will not be same ,then how will the pressure inside the container be balanced by smaller number of particles moving around with greater energy(so are at a higher temperature).
    But if i try to see this from the point of view of newton's third law, then the force between two particles is always equal and opposite on the two of them ... hence taking the system and surrounding as two huge particles...it seems they will never allow the pressure to differ, ensuring all the processes will be isobaric..
    But i am really confused about all this..
  5. Aug 30, 2012 #4
    Perhaps this will help you both.

    It is a good idea to understand the basics of thermo before learning the posh words like isobaric.

    So what are the basics?

    Well we need a system, we need the surroundings, we need the boundary between them and we need a process.

    A system is a part of the universe considered separate from the rest of the universe by a well defined boundary.

    The surroundings are the rest of the universe.

    The process describes the exchange of energy (heat, work or some other form) across the boundary between the system and the rest of the universe.

    That is the first law.

    It is vital to realise that the work or heat is transferred across the boundary and does not occur exclusively within the system or surroundings.

    This is important because we can calculate the heat transferred or the work done for either the surroundings or the system.

    Of course
    the heat transferred to the sytem = heat lost by the surroundings
    the work done by the system = the work done on the surroundings.

    So to isobaric processes.

    Because we can perform our calculation for either the system or the surroundings and get the same answer we calculate for the most convenient.

    We don't care what the pressure conditions within the system are or even if they are defined if we can say that the surroundings are at constant pressure. This is usually the case for chemical reactions or work done against the atmospheric pressure.

    So we can say that the heat transferred at constant pressure = Cp ΔTsystem when we heat a beaker of water in the atmosphere.

    If we heat a sealed container of gas ( balloon) in the atmosphere it expands against the pressure of the atmosphere ie the work done on the atmosphere (the surroundings) is done at constant pressure.
    Since the volume lost by the surroundings = the volume gain by the balloon we can say that the work done by the balloon on the surroundings = Patmosphere ΔV

    does this help?
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